HDU 3718 Similarity

题目链接

题意：给定一个标准答案字符串，然后下面每一行给一个串，要求把字符一种对应一种，要求匹配尽量多

思路：显然的KM最大匹配问题，位置对应的字符连边权值+1

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int MAXNODE = 27;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct KM {
	int n, m;
	Type g[MAXNODE][MAXNODE];
	Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];
	int left[MAXNODE], right[MAXNODE];
	bool S[MAXNODE], T[MAXNODE];

	void init(int n, int m) {
		this->n = n;
		this->m = m;
		memset(g, 0, sizeof(g));
	}

	void add_Edge(int u, int v, Type val) {
		g[u][v] += val;
	}

	bool dfs(int i) {
		S[i] = true;
		for (int j = 0; j < m; j++) {
			if (T[j]) continue;
			Type tmp = Lx[i] + Ly[j] - g[i][j];
			if (!tmp) {
				T[j] = true;
				if (left[j] == -1 || dfs(left[j])) {
					left[j] = i;
					right[i] = j;
					return true;
				}
			} else slack[j] = min(slack[j], tmp);
		}
		return false;
	}

	void update() {
		Type a = INF;
		for (int i = 0; i < m; i++)
			if (!T[i]) a = min(a, slack[i]);
		for (int i = 0; i < n; i++)
			if (S[i]) Lx[i] -= a;
		for (int i = 0; i < m; i++)
			if (T[i]) Ly[i] += a;
	}

	Type km() {
		memset(left, -1, sizeof(left));
		memset(right, -1, sizeof(right));
		memset(Ly, 0, sizeof(Ly));
		for (int i = 0; i < n; i++) {
			Lx[i] = -INF;
			for (int j = 0; j < m; j++)
				Lx[i] = max(Lx[i], g[i][j]);
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) slack[j] = INF;
			while (1) {
				memset(S, false, sizeof(S));
				memset(T, false, sizeof(T));
				if (dfs(i)) break;
				else update();
			}
		}
		Type ans = 0;
		for (int i = 0; i < n; i++) {
			ans += g[i][right[i]];
		}
		return ans;
	}
} gao;

const int N = 10005;

int t, n, k, m;
char s[N][2];

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d%d", &n, &k, &m);
		for (int i = 0; i < n; i++)
			scanf("%s", s[i]);
		while (m--) {
			gao.init(26, 26);
			char c[2];
			for (int i = 0; i < n; i++) {
				scanf("%s", c);
				gao.add_Edge(s[i][0] - 'A', c[0] - 'A', 1);
			}
			printf("%.4lf\n", gao.km() * 1.0 / n);
		}
	}
	return 0;
}